Tiêu đề 24. Atomic Physics -2 Easy Ans.pdf ✅

1. (a) For n=1, maximum number of states 2 2 2 = n = and for n = 2, 3, 4, maximum number of states would be 8, 18, 32 respectively, Hence number of possible elements = 2 + 8 + 18 + 32 = 60. 2. (d) Bohr radius 2 2 2 2 0 ; r n Zme n h r =     3. (a) E1→2 = −3.4 −(13.6) = +10.2 eV 4. (d)       = − 2 2 2 2 1 1 1 1 RZ  For di-ionised lithium the value of Z is maximum. 5. (c) Lyman series lies in the UV region. 6. (b) The size of the atom is of the order of 1Å = 10–10m. 7. (b) Balmer series lies in the visible region. 8. (c) Transition A (n =  to 1) : Series limit of Lyman series Transition B (n = 5 to n = 2) : Third spectral line of Balmer series Transition C (n = 5 to n = 3) : Second spectral line of Paschen series 9. (a) D is excitation of electron from 2nd orbit corresponding to absorption line in Balmer series and E is the energy released to bring the electron from  to ground state i.e. ionisation potential. 10. (d) 11. (b) Paschen series lies in the infrared region. 12. (b) Energy required to knock out the electron in the n th orbit eV n 2 13.6 = +  E eV 9 13.6 3 = + . 13. (b) Linear momentum 31 6 = = 9.1 10  2.2 10 − mv 2.0 10 kg m / s 24 =  − − 14. (c) 2 r  n  0 2 rn = n a ( ) 1 a0 r = 15. (c) For the ionization of second He electron. + He will act as hydrogen like atom. Hence ionization potential Z 13.6 volt (2) 13.6 54.4 V 2 2 =  =  = 16. (c) Energy required eV n 0.136 10 13.6 13.6 2 2 = = = 17. (c)         = − 2 2 2 1 1 1 1 n n R  n n R 1 1 1 2 2 2 1  − = 7 10 1.097 10 18752 10 1 −    = . 144 7 = 0.0486 = But 3 144 7 4 1 3 1 2 − 2 =  n1 = and n2 = 4 (Paschen series) 18. (b) Potential energy of electron in n th orbit of radius r in H- atom r e U 2 = − (in CGS) ∵ K.E. | . .| 2 1 = P E  r e K 2 2 = 19. (c) Final energy of electron = −13.6 +12.1 = −1.51 eV. which is corresponds to third level i.e. n = 3 . Hence number of spectral lines emitted 3 2 3(3 1) 2 ( 1) = − = − = n n 20. (b) Let the energy in A, B and C state be EA. EB and EC, then from the figure ( ) ( ) ( ) EC − EB + EB − EA = EC − EA or 1 2 3 hc hc hc + = 1 2 1 2 3      +  = 21. (c) According to Bohr’s second postulate. 22. (c) First excited state i.e. second orbit (n = 2) Second excited state i.e. third orbit (n = 3) 2 13.6 n  E = −  4 9 2 3 2 3 2  =      = E E 23. (c) cm R R R 5 2 2 10 3 16 3 16 16 3 4 1 2 1 1 −  =  = =       = −   n = 2 n = 1 E 3.4 eV (2) 13.6 2 2 = − = − E1 = −13.6 eV 1 2 3 C B A
Frequency Hz c n 1 5 5 1 0 10 16 9 10 3 16 3 10 =    = =  − 24. (d) Energy required to remove electron in the n = 2 state 3.4 eV (2) 13.6 2 = + = + 25. (d) (Eion)Na Z (Eion)H (11) 13.6 eV 2 2 = = 26. (c) The wavelength of spectral line in Balmer series is given by       = − 2 2 1 2 1 1 n R  For first line of Balmer series, n = 3  36 5 3 1 2 1 1 2 2 1 R R =       = −  ; For second line n = 4.  16 3 4 1 2 1 1 2 2 2 R R =      = −   6561 4860 Å 27 20 27 20 1 1 2 =   =  =   27. (d)   hc E hc 2E − E =  =       3 ' 3 ' 3 ' 3 ' 4 − =  =  =  = hc E hc E E 28. (b) Because atom is hollow and whole mass of atom is concentrated in a small centre called nucleus. 29. (d) Z n r Zme n h r 2 2 2 2 0 = ;     30. (b) 2 r  n  9 4 ( 3) ( 2) = = = n n r r  rn R 2.25 R 4 9 ( =3) = = 31. (a) In the revolution of electron, coulomb force provides the necessary centripetal force  r mv r ze 2 2 2 =  r ze mv 2 2 =  K.E. r ze mv 2 2 1 2 2 = = 32. (d) According to Bohr’s theory 2 h mvr = n  Circumference  n mv h r n  =      2 = 33. (c) r kZe K E 2 . 2 = and . . ; 2 r kZe P E = −  2 1 . . . . = − P E K E . 34. (d) Lyman series lies in the UV region. 35. (d) If E is the energy radiated in transition then ER→G  EQ→S  ER→S  EQ→R  EP→Q For getting blue line energy radiated should be maximum         1 E . Hence (d) is the correct option. 36. (b) Energy released 2.55 eV (4) 1 (2) 1 13.6 2 2  =      = − 37. (c) The absorption lines are obtained when the electron jumps from ground state (n = 1) to the higher energy states. Thus only 1, 2 and 3 lines will be obtained. 38. (a) P.E. r 1  − and K.E. r 1  As r increases so K.E. decreases but P.E. increases. 39. (c) Wave number 16 3 16 1 4 1 1 1 1 2 2 2 1 R R n n R =       = −         = −  40. (c) In hydrogen atom, the lowest orbit (n = 1) corresponds to minimum energy (– 13.6 eV). 41. (a) K.E. = – (T.E.) 42. (d) Required energy E 1.51eV 3 13.6 2 3 = + = 43. (d) As n increases P.E. also increases. 44. (a) When an electron jumps from the orbit of lower energy (n=1) to the orbit of higher energy (n=3), energy is absorbed. 45. (a) For Lyman series 4 3 (2) 1 (1) 1 2 2 max Lymen RC Rc c  =      = = −   For Balmer series 36 5 (3) 1 (2) 1 2 2 max Balmer RC Rc c  =      = = −    5 27 Balmer Lymen =   46. (a)  E1  E2 1  2 i.e. photons of higher frequency will be emitted if transition takes place from n = 2 to 1. r e + n=6 n=5 n=4 n=3 n=2 n=1 E2 E1

         = − 2 2 2 1 1 2 1 1 n n RZ  gives n n R n n Z ( ) 2 1 2 2 2 2 2 2 1 − = On putting values Z = 2 From eV n Z E 54.4 (1) 13.6 13.6(2) 2 2 2 2 = − − = − = 71. (a) Ionization energy = Binding energy. 72. (b) E = −Rch  8 34 19 3 10 6.6 10 13.6 1.6 10 − −      = − = − ch E R per m 7 = 1.098 10 73. (b) Bohr postulated that the angular momentum of the electron is conserved. 74. (d) 1.51 ; 9 13.6 3 E = − = − eV E 0.85 eV 16 13.6 4 = − = − E4 − E3 = 0.66 eV 75. (b) Number of spectral lines 6 2 4(4 1) 2 ( 1) = − = − = n n NE 76. (d) In the transition from orbit 5 → 2, more energy is liberated as compared to transition from 4 → 2. 77. (d) Impact parameter 2 cot  b  Here b = 0, hence o  = 180 78. (b) 2 2 . .          = i f i f n n r r r n i e  2 11 11 5.3 10 1 21.2 10       =   − − n  4 2 n =  n = 2 79. (a) 80. (d) eV n En 3.4 4 13.6 13.6 2 = − − = − = 81. (a) 36 5 3 1 2 1 1 2 2 Balmer R R =      = −  , 4 3 2 1 1 1 1 2 2 Lyman R R =      = −  1215 .4 Å 27 5 Lyman = Balmer  = 82. (a) . 1 1 1 2 2 2 1         = − n n RH  For Lyman series n1=1 and n2=2, 3, 4, When n2=2, we get cm RH 3 10967 4 3 4   = = 83. (b) 2 r  n . For ground state n=1 and for first excited state n=2. 84. (b) No. of lines 3 2 3(3 1) 2 ( 1) = − = − = n n NE 85. (d) Infinitely large transitions are possible (in principle) for the hydrogen atom. 86. (c) 2 rn  n 87. (a) 2 2 n hc Z E        =   2 1 Z   Hence cm He 5.099 4 20.397  + = = 88. (c) Excitation potential e Excitation energy = Minimum excitation energy corresponds to excitation from n =1 to n = 2  Minimum excitation energy in hydrogen atom = −3.4 −(−13.6) = +10.2 eV so minimum excitation potential = 10.2 eV. 89. (a) Orbital speed varies inversely as the radius of the orbit. Energy increases with the increase in quantum number. 90. (b)         = − 2 2 2 1 1 1 1 n n R   36 5 (3) 1 (2) 1 1 2 2 3 2 R R  =      = −  → and 16 3 (4) 1 (2) 1 1 2 2 4 2 R R  =      = −  →  27 20 3 2 4 2 = → →    4 2 0 27 20  → =  91. (d) 92. (b) Kinetic energy = |Total energy| 93. (c) Energy to excite the − e from n =1 to n = 2 1 9 3.4 ( 13.6) 10.2 10.2 1.6 10 − E = − − − = eV =   J 18 1.632 10 − =  94. (c) First excited state Ground state n = 1 (–13.6 eV) n = 2 (– 3.4 eV) (For H2 - atom)